Intersection Points & Finding Unknown Variable

[confusedatmath]

The line with equation y = x + k, where k is a real number, intersects the parabola with equation y = x^2 + x − 2 in two distinct points if

I first made the equations equal each other

x + k = x^2 + x − 2
0 = x^2 -2 -k

From here i thought you use the discriminate a=1 b=o c=-2-k

but this isn't right, because the answer to choose from

k < − 2

k > − 2

k = − 2

k < 2

k ≠ 2

 

[mente oscura]

The line with equation y = x + k, where k is a real number, intersects the parabola with equation y = + x − 2 in two distinct points if

I first made the equations equal each other

x + k = + x − 2
0 = -2 -k

From here i thought you use the discriminate a=1 b=o c=-2-k

but this isn't right, because the answer to choose from

k < − 2

k > − 2

k = − 2

k < 2

k ≠ 2

Hello.

Check the wording of the question. The parable, need that 'x' is not high to the power 1. (would be a straight line)

Regards.

 

[confusedatmath]

Hello.

Check the wording of the question. The parable, need that 'x' is not high to the power 1. (would be a straight line)

Regards.

i fixed it :p read again, it was a mistake i forgot the x^2

 

[mente oscura]

The line with equation y = x + k, where k is a real number, intersects the parabola with equation y = x^2 + x − 2 in two distinct points if

I first made the equations equal each other

x + k = x^2 + x − 2
0 = x^2 -2 -k

From here i thought you use the discriminate a=1 b=o c=-2-k

but this isn't right, because the answer to choose from

k < − 2

k > − 2

k = − 2

k < 2

k ≠ 2

Now yes.

0=x^2-2-k

x^2=k+2

x=\pm{} \sqrt{k+2}

1º) k&lt;-2 \rightarrow{}x \cancel{\in}{R}

2º) k&gt;-2 \rightarrow{}x \in{R}

3º) k=-2 \rightarrow{}x=0, only a breakpoint.

Regards.